The $\displaystyle x$-intercept is $\displaystyle (a,0)$, where $\displaystyle f(a)= 0$:

$\displaystyle f(x) = 5 \cdot 4^{x- 3}- 3$

$\displaystyle 5 \cdot 4^{a- 3}- 3 = 0$

$\displaystyle 5 \cdot 4^{a- 3}- 3 + 3= 0 + 3$

$\displaystyle 5 \cdot 4^{a- 3}= 3$

$\displaystyle 5 \cdot 4^{a- 3} \div 5= 3 \div 5$

$\displaystyle 4^{a- 3} = 0.6$

$\displaystyle \ln \left (4^{a- 3} \right )=\ln 0.6$

$\displaystyle (a- 3)\ln 4 =\ln 0.6$

$\displaystyle a- 3=\frac{\ln 0.6}{\ln 4 } \approx \frac{-0.5108}{1.3863}\approx -0.3685$

$\displaystyle a- 3+3 \approx -0.3685 + 3$

$\displaystyle a \approx 2.63$

The $\displaystyle x$-intercept is $\displaystyle (2.63,0)$.

The $\displaystyle y$-intercept is $\displaystyle (0, f(0))$:

$\displaystyle f(x) = 5 \cdot 4^{x- 3}- 3$

$\displaystyle f(0) = 5 \cdot 4^{0- 3}- 3$

$\displaystyle f(0) = 5 \cdot 4^{ - 3}- 3$

$\displaystyle f(0) = 5 \cdot \frac{1}{64}- 3$

$\displaystyle f(0) = \frac{5}{64}- 3$

$\displaystyle f(0) \approx 0.08 - 3 \approx -2.92$

$\displaystyle (0, -2.92)$ is the $\displaystyle y$-intercept.

Since 4 can be raised to the power of any real number, the domain of $\displaystyle f$ is the set of all real numbers. Therefore, there is no vertical asymptote of the graph of $\displaystyle f$.

We can rewrite this as follows:

$\displaystyle g(x) = 9 \cdot 3^{x}- 3$

$\displaystyle g(x) = 3^{2} \cdot 3^{x}- 3$

$\displaystyle g(x) = 3^{x+2}- 3$

This is a translation of the graph of $\displaystyle f(x) = 3^{x}$, which has $\displaystyle y = 0$ as its horizontal asymptote, to the right two units and down three units. Because of the latter translation, the horizontal asymptote is $\displaystyle y = -3$.

We can rewrite the statements using $\displaystyle y$ for both $\displaystyle f(x)$ and $\displaystyle g(x)$ as follows:

$\displaystyle y = e^{x}+ 9$

$\displaystyle y = 4e^{x}- 7$

To solve this, we can multiply the first equation by $\displaystyle -4$, then add:

$\displaystyle -4y = -4e^{x}-36$

      $\displaystyle \underline{y = 4e^{x} \; \; - 7}$

$\displaystyle -3y =$            $\displaystyle -43$

$\displaystyle -3y \div (-3) = -43 \div (-3 )$

$\displaystyle y \approx 14.3$

We can rewrite the statements using $\displaystyle y$ for both $\displaystyle f(x)$ and $\displaystyle g(x)$ as follows:

$\displaystyle y = e^{x}+ 9$

$\displaystyle y = 4e^{x}- 7$

To solve this, we can set the expressions equal, as follows:

$\displaystyle 4e^{x}- 7 = e^{x}+ 9$

$\displaystyle 4e^{x}- 7 - e^{x} + 7 = e^{x}+ 9 - e^{x} + 7$

$\displaystyle 3e^{x} = 16$

$\displaystyle 3e^{x}\div 3 = 16 \div 3$

$\displaystyle e^{x} = \approc 5.3333$

$\displaystyle x \approx \ln 5.333 \approx 1.7$

$\displaystyle The\ graph\ of\ the\ function\ y=3^{x}-2\ has\ an\ asymptote\ at\ y=-2.$

$\displaystyle An\ asymptote\ on\ a\ graph\ is\ a\ value\ that\ a \ function\ approaches\ but\ never\ touches.$

$\displaystyle Since\ this\ graph\ approaches\ but\ never\ touches\ -2\ we\ use\ the >\ symbol.$

$\displaystyle \\Also\ because\ the\ 3\ is\ positive\ that\ means\ the\ graph\ is\ concave\ up\ so\ >\ is\ chosen\ instead\ of\ < .$

$\displaystyle \\The\ concavity\ of\ an\ exponential\ graph\ can\ be\ determined\ by\ its\ a-value.$

$\displaystyle This\ function\ is\ in\ the\ form\ y=a(b)^{x}+k.$

$\displaystyle \ So\ the\ a\ value\ is\ 2, based\ on\ the\ function\ we\ were\ given\ y=2(3)^{x}-5$

$\displaystyle If\ the\ a\ value\ is\ positive,\ the\ graph\ will\ be\ concave\ up$.

$\displaystyle If\ the\ a\ value\ is\ negative,\ the\ graph\ will\ be\ concave\ down.$

$\displaystyle To\ find\ the\ y-intercept,\ plug\ in\ x=0, and\ keep\ in\ mind\ order\ of\ operations,$

$\displaystyle \ Exponent first, then\ multiply.$

$\displaystyle y=2(3)^{0}-5$

$\displaystyle y=2(1)-5=-3$

$\displaystyle The\ y-intercept\ is\ (0,-3).$

Define $\displaystyle g(x) = e^{x}$. In terms of $\displaystyle g(x)$, $\displaystyle f(x)$ can be restated as

$\displaystyle f(x) = 2g( x-3 )+ 7$

The graph of $\displaystyle f(x)$ is a transformation of that of $\displaystyle g(x)$. As an exponential function, $\displaystyle g(x)$ has a graph that has no vertical asymptote, as $\displaystyle g(x)$ is defined for all real values of $\displaystyle x$; it follows that being a transformation of this function, $\displaystyle f(x)$ also has a graph with no vertical asymptote as well.

Define $\displaystyle g(x) = e ^{x}$. In terms of $\displaystyle g(x)$, $\displaystyle f(x)$ can be restated as

$\displaystyle f(x) = 2e^{x-3}- 7$.

The graph of $\displaystyle g(x)$ has as its horizontal asymptote the line of the equation $\displaystyle y = 0$. The graph of $\displaystyle f(x)$ is a transformation of that of $\displaystyle g(x)$ - a right shift of 3 units ( $\displaystyle x-3$ ), a vertical stretch ( $\displaystyle 2g$ ), and a downward shift of 7 units ( $\displaystyle -7$ ). The right shift and the vertical stretch do not affect the position of the horizontal asymptote, but the downward shift moves the asymptote to the line of the equation $\displaystyle y = -7$. This is the correct response.